保持简单：

// Generator
function* iter(a, b, step = 1) {
  for (let i = b ? a : 0; i < (b || a); i += step) {
    yield i
  }
}

const range = (a, b, step = 1) =>
  typeof a === 'string'
    ? [...iter(a.charCodeAt(), b.charCodeAt() + 1)].map(n => String.fromCharCode(n))
    : [...iter(a, b, step)]

range(4) // [0, 1, 2, 3]
range(1, 4) // [1, 2, 3]
range(2, 20, 3) // [2, 5, 8, 11, 14, 17]
range('A', 'C') // ['A', 'B', 'C']

您还可以执行以下操作：

const range = Array.from(Array(size)).map((el, idx) => idx+1).slice(begin, end);

如果我们输入类似[4,2]的内容，我们将得到[2,3,4]作为输出，我们可以使用它。

function createRange(array) {
  var range = [];
  var highest = array.reduce(function(a, b) {
    return Math.max(a, b);
  });
  var lowest = array.reduce(function(a, b) {
    return Math.min(a, b);
  });
  for (var i = lowest; i <= highest; i++) {
    range.push(i);
  }
  return range;
}

要紧密复制的类型脚本函数

/**
 * Create a generator from 0 to stop, useful for iteration. Similar to range in Python.
 * See: https://stackoverflow.com/questions/3895478/does-javascript-have-a-method-like-range-to-generate-a-range-within-the-supp
 * See: https://docs.python.org/3/library/stdtypes.html#ranges
 * @param {number | BigNumber} stop
 * @returns {Iterable<number>}
 */
export function range(stop: number | BigNumber): Iterable<number>
/**
 * Create a generator from start to stop, useful for iteration. Similar to range in Python.
 * See: https://stackoverflow.com/questions/3895478/does-javascript-have-a-method-like-range-to-generate-a-range-within-the-supp
 * See: https://docs.python.org/3/library/stdtypes.html#ranges
 * @param {number | BigNumber} start
 * @param {number | BigNumber} stop
 * @returns {Iterable<number>}
 */
export function range(
  start: number | BigNumber,
  stop: number | BigNumber,
): Iterable<number>

/**
 * Create a generator from start to stop while skipping every step, useful for iteration. Similar to range in Python.
 * See: https://stackoverflow.com/questions/3895478/does-javascript-have-a-method-like-range-to-generate-a-range-within-the-supp
 * See: https://docs.python.org/3/library/stdtypes.html#ranges
 * @param {number | BigNumber} start
 * @param {number | BigNumber} stop
 * @param {number | BigNumber} step
 * @returns {Iterable<number>}
 */
export function range(
  start: number | BigNumber,
  stop: number | BigNumber,
  step: number | BigNumber,
): Iterable<number>
export function* range(a: unknown, b?: unknown, c?: unknown): Iterable<number> {
  const getNumber = (val: unknown): number =>
    typeof val === 'number' ? val : (val as BigNumber).toNumber()
  const getStart = () => (b === undefined ? 0 : getNumber(a))
  const getStop = () => (b === undefined ? getNumber(a) : getNumber(b))
  const getStep = () => (c === undefined ? 1 : getNumber(c))

  for (let i = getStart(); i < getStop(); i += getStep()) {
    yield i
  }
}

它适用于字符和数字，通过可选步骤向前或向后移动。

var range = function(start, end, step) {
    var range = [];
    var typeofStart = typeof start;
    var typeofEnd = typeof end;

    if (step === 0) {
        throw TypeError("Step cannot be zero.");
    }

    if (typeofStart == "undefined" || typeofEnd == "undefined") {
        throw TypeError("Must pass start and end arguments.");
    } else if (typeofStart != typeofEnd) {
        throw TypeError("Start and end arguments must be of same type.");
    }

    typeof step == "undefined" && (step = 1);

    if (end < start) {
        step = -step;
    }

    if (typeofStart == "number") {

        while (step > 0 ? end >= start : end <= start) {
            range.push(start);
            start += step;
        }

    } else if (typeofStart == "string") {

        if (start.length != 1 || end.length != 1) {
            throw TypeError("Only strings with one character are supported.");
        }

        start = start.charCodeAt(0);
        end = end.charCodeAt(0);

        while (step > 0 ? end >= start : end <= start) {
            range.push(String.fromCharCode(start));
            start += step;
        }

    } else {
        throw TypeError("Only string and number types are supported");
    }

    return range;

}

jsFiddle。

如果扩充本机类型是您的事情，那么将其分配给Array.range。

var范围=函数（开始、结束、步骤）{var范围=[]；var typeofStart=启动类型；var typeofEnd=结束类型；如果（步骤==0）{throw TypeError（“步长不能为零。”）；}if（类型开始==“undefined”| |类型结束==“未定义”）{throw TypeError（“必须传递开始和结束参数。”）；}否则如果（typeofStart！=typeofEnd）{throw TypeError（“开始和结束参数必须是相同的类型。”）；}步骤类型==“未定义”&&（步骤=1）；if（结束<开始）{step=-步骤；}if（开始类型==“number”）{while（步骤>0？结束>=开始：结束<=开始）{范围.推（启动）；开始+=步骤；}}否则if（typeofStart==“string”）{如果（start.length！=1 | | end.length；=1）{throw TypeError（“仅支持带有一个字符的字符串。”）；}start=start.charCodeAt（0）；end=end.charCodeAt（0）；while（步骤>0？结束>=开始：结束<=开始）{range.push（String.fromCharCode（开始））；开始+=步骤；}}其他{throw TypeError（“仅支持字符串和数字类型”）；}返回范围；}console.log（范围（“A”，“Z”，1））；console.log（范围（“Z”，“A”，1））；console.log（范围（“A”，“Z”，3））；console.log（范围（0，25，1））；console.log（范围（0，25，5））；console.log（范围（20，5，5））；

没有一个示例进行了测试，每个步骤都有一个生成递减值的选项。

export function range(start = 0, end = 0, step = 1) {
    if (start === end || step === 0) {
        return [];
    }

    const diff = Math.abs(end - start);
    const length = Math.ceil(diff / step);

    return start > end
        ? Array.from({length}, (value, key) => start - key * step)
        : Array.from({length}, (value, key) => start + key * step);

}

测验：

import range from './range'

describe('Range', () => {
    it('default', () => {
        expect(range()).toMatchObject([]);
    })

    it('same values', () => {
        expect(range(1,1)).toMatchObject([]);
    })

    it('step=0', () => {
        expect(range(0,1,0)).toMatchObject([]);
    })

    describe('step=1', () => {
        it('normal', () => {
            expect(range(6,12)).toMatchObject([6, 7, 8, 9, 10, 11]);
        })

        it('reversed', () => {
            expect(range(12,6)).toMatchObject([12, 11, 10, 9, 8, 7]);
        })
    })

    describe('step=5', () => {

        it('start 0 end 60', () => {
            expect(range(0, 60, 5)).toMatchObject([0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55]);
        })

        it('reversed start 60 end -1', () => {
            expect(range(55, -1, 5)).toMatchObject([55, 50, 45, 40, 35, 30, 25, 20, 15, 10, 5, 0]);
        })
    })
})